From the definition of Gibbs free energy, we can write for an irreversible process:


With $TdS=\pu{q_{rev}//T}$ this becomes:


Since G is defined for constant (p,T) $dH=dq$, in the case of an irreversible process we can write $dq=dq_{irrev}$, so:


Or, by default:


BUT: since $dH=dq$, this would mean:


Which is in contradiction with the fact that H is a state function (the path between 2 states shouldn't matter)! What am I missing here?

  • $\begingroup$ When you write dG=dH-TdS, you are implying that T is constant. Did you mean to imply this? Can you give a specific process example to focus on to illustrate what you are asking? $\endgroup$ Commented Dec 11, 2018 at 14:55
  • $\begingroup$ The definition of Gibbs free energy always implies constant T. It is valid for all processes assuming constant T and p. $\endgroup$
    – Stikke
    Commented Dec 11, 2018 at 21:30
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    $\begingroup$ This is incorrect. The definition of Gibbs free energy is G=H-TS, so dG=dH-TdS-SdT=VdP-SdT $\endgroup$ Commented Dec 11, 2018 at 21:35
  • $\begingroup$ Can whoever downvoted this question tell me why? Or answer the question at least? I asked this to quite some people with knowledge of thermodynamics and they weren't able to answer me so it must be a valid question. $\endgroup$
    – Stikke
    Commented Dec 11, 2018 at 21:36
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    $\begingroup$ It wasn't me. I will try to help you. $\endgroup$ Commented Dec 11, 2018 at 21:37

3 Answers 3


Even though the OP does not seem to be active in this tread any longer, I'm going to provide an answer anyway, since it seems like such an interesting question.

First of all, even if a closed system undergoes a process in which it is in contact with a constant pressure surroundings throughout the entire process, and this pressure is the same as that of the system in its initial and final states, the heat absorbed Q is still not necessarily equal to the change in enthalpy unless PV work is the only form of work being done. For example, if there is a stirrer doing work to agitate the system (and cause irreversible viscous heating), and no PV work occurs, and the system is in contact with a constant temperature bath throughout, the change in enthalpy will be equal to zero, but Q will be equal to minus the amount of work that the stirrer does on the system.

Secondly, assume that throughout the irreversible process, the system undergoes a change in which it is in contact with a single constant temperature reservoir at the same temperature as the initial and final temperatures of the system, and is also in contact with a constant pressure surroundings at the same pressure as that of the system in its initial and final states. But consider that a reversible process between the same initial and final end states does not have to resemble the irreversible path in any way whatsoever, as long as it matches the initial and final temperatures and pressures (and species concentrations if chemical reaction is involved). Even if the boundary temperature for the reversible path is held constant at the same value as the irreversible path, the pressure of the surroundings during the reversible path definitely does not have to be (and will not be) constant throughout the reversible path. And, given that this is the case, the heat Q for the reversible path would not be required to be the same as that for the irreversible path, even if the enthalpy changes for both paths is the same. In virtually all cases, it will actually not be possible to devise a reversible path at a constant surrounding pressure between the same pair of end states. So the heat Q for the reversible path will not be equal to the change in enthalpy for the reversible path (and thus the change in enthalpy for the irreversible path).

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    $\begingroup$ Thanks and sorry for my inactivity first of all. Regarding the first point, ok true but let's assume only pV work. In the second point you make the case that qrev is not equal to q(irrev), which I agree with. But since H=pV+U=q-pdV+Vdp+pdV=q+Vdp, H equals q at constant pressure in absence of pV work. So saying q(rev) is not equal to q(irrev) is saying H(rev) is not H(irrev). And H is a state function (unlike q). That was the origin of my confusion. $\endgroup$
    – Stikke
    Commented Dec 26, 2018 at 10:22
  • $\begingroup$ Please don't use differentials for finite path lengths in irreversible processes. It will lead you astray. Now, let's focus on a specific irreversible process, and see how this plays out. I am proposing we look at the irreversible expansion of an ideal gas, in which the gas is held in contact with a constant temperature reservoir at its initial temperature and in which we suddenly (at time zero) drop the external pressure to a new constant value and then hold it at that value until the gas re-equilibrates. Is this an OK case to consider? $\endgroup$ Commented Dec 26, 2018 at 12:56
  • $\begingroup$ The problem is that H is defined for constant pressure, usually H is used for processes involving a change in chemical potential. I guess an isobaric expansion would fit the definition of H (although for G as stated in the original question both T and p are constant) $\endgroup$
    – Stikke
    Commented Dec 26, 2018 at 13:09
  • $\begingroup$ H is not defined for constant pressure. The definition of H is simply H = U+PV. Are you saying you don't like the specific focal problem I proposed? If not, it's your turn to propose a focal problem. $\endgroup$ Commented Dec 26, 2018 at 13:47
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Commented Dec 27, 2018 at 2:51

The simple answer to your question is that, for an irreversible process, $\text{d}H \neq \text{d}q_\text{rev}$ under the given constraints (i.e., constant $T$ and $P$, $P_\text{sys} = P_\text{surr}$, only $PV$ work). On the other hand, $\text{d}H = \text{d}q_\text{irrev}$ is valid. As Chet Miller suggests in his answer, there is no reversible path for a change of composition at constant $T$ and $P$ (when there is only $PV$ work). Thus, while $\text{d}H = \text{d}q$ at constant $T$ and $P$, $\text{d}H$ nevertheless cannot equal $\text{d}q_\text{rev}$, simply because $\text{d}q_\text{rev}$ does not exist at constant $T$ and $P$.

Note that at equilibrium, there is no change of composition, $\text{d}G=0$ and $\text{d}H = \text{d}q = \text{d}q_\text{rev}$ as it should.

One way to have a reversible change of composition at constant $T$ and $P$ is to allow non-$PV$ work, such as that in an electrochemical cell. By carefully and slowly adjusting the current, we can allow the reaction to proceed reversibly. However, in that case $\text{d}H \neq \text{d}q$. In the presence of non-$PV$ work, $\text{d}H = \text{d}q + \text{d}w_o$. If the non-$PV$ work allows the reaction to go reversibly, then:

\begin{align} \text{d}H = \text{d}q_\text{rev} + \text{d}w_o \end{align}


\begin{align} \text{d}G &= \text{d}H - T \text{d}S \\ &= \text{d}q_\text{rev} + \text{d}w_o - T \text{d}S \\ &= \text{d}w_o \end{align}


In your last line, you claim that $\text{d}H_\text{irrev} = \text{d}q_\text{irrev}$. But this cannot be true, because $\text{d}q_\text{irrev}$ is an inexact differential, and it does not then make sense to talk about $\text{d}H_\text{irrev}$ because the system’s thermodynamic state is ill-defined along irreversible paths. $\text{d}H = \text{d}q|_\text{const. P}$ only for reversible isobaric processes.

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    $\begingroup$ Suppose I have an ideal gas in a cylinder at temperature $T_0$ and pressure P. The external pressure on the piston is held at P throughout the process. I suddenly place the cylinder in contact with a bath at temperature $T_1$ and hold it in contact with the bath until the system re-equilibrates. In this irreversible process, the work is $P\Delta V$ and, the heat added is equal to $Q=\Delta H$. Now, same setup, except here I add heat to the system very gradually using a sequence of reservoirs and temperatures running from $T_0$ and $T_1$. In this reversible path, $Q=\Delta H$ again. So??? $\endgroup$ Commented Dec 12, 2018 at 3:10
  • $\begingroup$ @ChesterMiller, your example involves states and finite differences $\Delta$. My answer concerns paths and infinitesimal differentials $d$. $\endgroup$ Commented Dec 12, 2018 at 4:06
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    $\begingroup$ How can you consider differential changes in a state function like H along irreversible paths if the states along the path are not thermodynamic equilibrium states? $\endgroup$ Commented Dec 12, 2018 at 4:15
  • $\begingroup$ @ChesterMiller, that’s a good point, and I have updated my answer accordingly. The rest of my answer and my response to your proposed counterexample should still be valid, however. $\endgroup$ Commented Dec 12, 2018 at 4:25
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    $\begingroup$ I do not understand this. The derivation I've seen is exactly this: H=U+pV, so dH=dq-pdV+pdV+Vdp. So at constant pressure and temperature this becomes dH=dq. Enthalpy is used a lot in chemistry, and is always treated as a state function (also Wikipedia confirms btw :-)), depending only on initial and final state. Just like U is a state function and Delta(U) is the same for reversible and irreversible processes H should be as well. You are then basically saying that H is not a state function. $\endgroup$
    – Stikke
    Commented Dec 26, 2018 at 10:31

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